Bus Timetable: A Quadratic, Two Roots and Two Rival Claims
A bus covers 240 km between two towns. On ordinary days it maintains an average speed of x km/h and takes the usual time. On a festival day, heavy traffic forces the average speed down by 8 km/h, and the journey takes exactly 1 hour longer. The depot manager writes an equation and obtains two values, 48 and −40, then tells the conductor to use 48 because it is positive. The conductor wonders whether the negative value should simply be discarded or whether it carries meaning. A second driver claims that on a clear day, raising the ordinary speed by 10 km/h would shorten the 240 km trip by exactly 1 hour. Before the timetable is reprinted, the depot needs to know which claim is consistent with the data and how the two roots should be interpreted.
1. Form the equation for the festival-day trip, solve it, and interpret both roots.
Read the worked answer
Let the ordinary speed be x km/h. The ordinary time is 240/x hours. The festival speed is (x − 8) km/h, so the festival time is 240/(x − 8) hours. This is 1 hour more:
240/(x − 8) = 240/x + 1.
Multiply both sides by x(x − 8), noting x ≠ 0 and x ≠ 8:
240x = 240(x − 8) + x(x − 8) 240x = 240x − 1920 + x² − 8x 0 = x² − 8x − 1920.
The discriminant is (−8)² − 4(1)(−1920) = 64 + 7680 = 7744 = 88². Thus
x = (8 ± 88)/2 = 48 or −40.
The root 48 gives ordinary time 240/48 = 5 hours and festival time 240/40 = 6 hours, satisfying the 1-hour increase. The root −40 does satisfy the multiplied equation but represents a negative speed, which is meaningless here; it is an algebraic artefact of clearing denominators. The physical answer is x = 48 km/h. Discarding −40 is justified by the physical domain, not merely because it is negative.
2. Test the second driver's claim that a 10 km/h increase saves exactly 1 hour. What increase would actually save 1 hour?
Read the worked answer
At the ordinary speed of 48 km/h, the ordinary time is 240/48 = 5 hours.
If the speed is raised by 10 km/h to 58 km/h, the time becomes 240/58 ≈ 4.138 hours. The saving is about 5 − 4.138 = 0.862 hours, roughly 52 minutes, not 1 hour. So the second driver's claim is false.
To find the increase y that saves exactly 1 hour, require
240/(48 + y) = 5 − 1 = 4.
Then 48 + y = 240/4 = 60, so y = 12 km/h. A 12 km/h increase, to 60 km/h, is needed; a 10 km/h increase falls short. The depot should reprint the timetable using 48 km/h for ordinary days, 40 km/h for the festival-day allowance, and should not accept the 10 km/h claim without calculation.